How do you find #int (7x^2+x-1) / ((2x+1) (x^2-4x+4)) dx# using partial fractions?

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Answer 1 · 2015-11-15T21:42:38

#I=1/50ln|2x+1|+87/25ln|x-2|-29/(5(x-2))+C#

#(x^2-4x+4)=(x-2)^2#

#(7x^2+x-1)/((2x+1)(x-2)^2)=A/(2x+1)+B/(x-2)+C/(x-2)^2=#

#=(A(x-2)^2+B(2x+1)(x-2)+C(2x+1))/((2x+1)(x-2)^2)=#

#=(A(x^2-4x+4)+B(2x^2-3x-2)+C(2x+1))/((2x+1)(x-2)^2)=#

#=(Ax^2-4Ax+4A+2Bx^2-3Bx-2B+2Cx+C)/((2x+1)(x-2)^2)=#

#=(x^2(A+2B)+x(-4A-3B+2C)+(4A-2B+C))/((2x+1)(x-2)^2)#

#A+2B=7 => A=7-2B#
#-4A-3B+2C=1#
#4A-2B+C=-1#

#-28+8B-3B+2C=1#
#28-8B-2B+C=-1#

#5B+2C=29#
#-10B+C=-29#

#5C=29 => C=29/5#

#B=(29-2C)/5=29/5-58/25=87/25#

#A=7-2B = 7-174/25=1/25#

#I=int (7x^2+x-1)/((2x+1)(x-2)^2) dx#

#I=1/25int dx/(2x+1)+87/25int dx/(x-2)+29/5intdx/(x-2)^2#

#I=1/50ln|2x+1|+87/25ln|x-2|-29/(5(x-2))+C#